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(avs294) seunsom HW7 bohm (59970) This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A bucket full of water is rotated in a vertical circle of radius 0.96 m (the approximate length of a persons arm). What must be the minimum speed of the pail at the top of the circle if no water is to spill out? Correct answer: 3.06725 m/s. Explanation: If we analyze the forces on the water at the top of a vertical circle (using the notation Fji for a two body force on from we see that i j), v2 . r To minimize the velocity, we minimize the lhs of the equation. Since we cant change the weight of the water, we use the lowest Normal force that we can, 0. Then, =m v2 mg = m r and so, v= = gr (9.8 m/s2 ) (0.96 m) Npail,water + Wearth,water = m acentripetal N = mg m N =m v2 r v2 g r
1
.
N = (2880 kg)
(14.6 m/s)2 9.8 m/s 60.3 m = 18043.2 N .
2
003 (part 2 of 2) 10.0 points What is the maximum speed the car can have as it passes this highest point before losing contact with the road? Correct answer: 24.3093 m/s. Explanation: When the car just loses contact with the road, the normal force becomes zero. At this point, mv 2 r v = gr =
mg =
(9.8 m/s2 )(60.3 m)
= 24.3093 m/s .
= 3.06725 m/s . 002 (part 1 of 2) 10.0 points An 2880 kg car passes over a bump in a road that follows the arc of a circle of radius 60.3 m. The acceleration of gravity is 9.8 m/s2 . What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 14.6 m/s? Correct answer: 18043.2 N. Explanation: At the highest point of the bump, Newtons second law in the vertical direction provides Fc = mg N
004 10.0 points An earth satellite remains in orbit at a distance of 5410 km from the center of the earth. What speed would it have to maintain? The universal gravitational constant is 6.67 1011 N m2 /kg2 and the mass of the earth is 5.98 1024 kg. Correct answer: 8586.47 m/s. Explanation:
Let;
r = 5410 km = 5.41 106 m , G = 6.67 1011 N m2 /kg2 , M = 5.98 1024 kg .
and
seunsom (avs294) HW7 bohm (59970) The gravitational force supplies the centripetal force, so GM m m v2 = r2 r GM v= r = 6.67 1011 N m2 /kg2 5.41 106 m 5.98 1024 kg
2
A Ferris wheel rotates at a constant rate. A 150 lb student has an apparent weight of 125 lb at the top. What is her apparent weight at the lowest point? 1. 175 lb correct 2. 225 lb 3. 125 lb 4. 100 lb 5. 150 lb 6. 200 lb
= 8586.47 m/s . 005
10.0 points
R
Explanation:
Let : Consider a satellite moving near the Earth surface, the radius of its orbit r is approximately the radius of the Earth R. The period T of the satellite is 1. T = R g R correct g
W = 150 lb and WT = 125 lb .
At the top, v2 WT = m . R At the bottom, WB W = m v2 R
2. T = 2 3. T = g R
WB = W + m
g R Explanation: 4. T = 2 g = 2 R , so g , and R 2 , so T= = = 2 006 R g
v2 R = W + (W WT ) = 2 W WT = 2 (150 lb) 125 lb = 175 lb .
10.0 points
007 10.0 points In a distant system, solar a giant planet has a small moon moving in a circular orbit of radius 4.99 105 km with an orbital period of 61.1 hrs. Find the mass of the planet. You will need G = 6.67 1011 m3 /kg/s2 . Correct answer: 1.52002 1027 kg.
seunsom (avs294) HW7 bohm (59970) Explanation: The centripetal force of the moon is m v2 Fc = = m 2 r , r and the gravitational force acting on the moon by the planet is Fg = GM m . r2 It is also true that R = re + rm Mm = re + re Me R re = Mm 1+ Me 3.84 108 m = 7.36 1022 kg 1+ 5.98 1024 kg = 3.45653 108 m .
3
Since the two forces should be equal, we have M= r3 2 G (4.99 108 m)3 (2.85651 105 s1 )2 = 6.67 1011 m3 /kg/s2 = 1.52002 1027 kg .
008 (part 1 of 2) 10.0 points On the way to the moon, the Apollo astronauts reach a point where the Moons gravitational pull is stronger than that of Earths. Find the distance of this point from the center of the Earth. The masses of the Earth and the Moon are 5.98 1024 kg and 7.36 1022 kg, respectively, and the distance from the Earth to the Moon is 3.84 108 m. Correct answer: 3.45653 108 m. Explanation: Let :
009 (part 2 of 2) 10.0 points What is the acceleration due to the Earths gravity at this point? The value of the universal gravitational constant is 6.672 1011 N m2 /kg2 . Correct answer: 0.00333946 m/s2 . Explanation: F G Me = 2 m re a = (6.672 1011 N m2 /kg2 ) 5.98 1024 kg (3.45653 108 m)2 a= = 0.00333946 m/s2 .
Me = 5.98 1024 kg , mm = 7.36 1022 kg , Re = 3.84 108 m .
and
Consider the point where the two forces are equal. If re is the distance from this point to the center of the Earth and rm is the distance from this point to the center of the Moon, then G m Me G m Mm = 2 2 re rm 2 Mm rm = 2 re Me rm = re Mm . Me
010 10.0 points If our Sun were twelve times as massive as it is, how many times faster or slower should the Earth move in order to remain in the same orbit? Correct answer: 3.4641. Explanation: Let : M = 4 M .
The gravitational force supplies the centripetal force: GM m m v2 = . F= r r2
seunsom (avs294) HW7 bohm (59970) v= GM M for the same distance, so r v 12 M = 12 = 3.4641 . = v M v M tells us that more massive bodies have higher speeds for a given orbit. 011 10.0 points Two satellites A and B orbit the Earth in the same plane. Their masses are 4 m and 5 m, respectively, and their radii 3 r and 5 r, respectively.
4
vB 25 = vA 12 Explanation: The force of gravity is responsible for holding a satellite in its orbit, so the orbital centripetal force is equal to the force of gravity: 9. Fr = m v2 ME m =G , r r2
where ME is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit (from the Earths center). Thus the tangential speed v of an orbit at distance r is v= GM r 1 . r
5r 3r A B 4m 5m
Since
3r 3 rA = = , the ratio rB 5r 5 vB = vA rA = rB 3 . 5
What is the ratio of their orbital speeds? 1. 2. 3. 4. 5. 6. vB = vA vB = vA vB = vA vB = vA vB = vA vB = vA 4 5 5 3 5 4 12 25 3 4 3 correct 5
7. None of these 8. vB = vA 4 3

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